chung to :1/41+1/42+1/43+.....+1/80>7/12
chung to A bang 1/41+1/42+1/43+...+1/80>7/12
S=1/41+1/42+1/43+.....+1/80
Chung to S>7/12
A=1\41+1\42+1\43+...+1\79+1\80.chung minh A >7\12
CMR 1/41+1/42+1/43+...+1/79+7/80 <7/12
Chứng minh:7/12< 1/41+1/42+1/43+...+1/79+1/80<1
A<10(1/40+1/50+1/70+1/60)=319/420<1
A>10(1/50+1/60+1/70+1/80)>7/12
=>7/12<A<1
cmr;M=1/41+1/42+1/43+..+1/80>7/12
CM:1/41 + 1/42+1/43+...+1/79+1/80 >7/12
Chứng tỏ: 1/41+1/42+1/43+.........+1/80 > 7/12
Đặt 1/41 + 1/42 + .... + 1/60 ( có 20 phân số )
1/61 + 1/62 + .... + 1/80 ( có 20 phân số )
Ta có : 1/41 + 1/42 + .... + 1/60 > 1/60 + 1/60 + .... + 1/60 = 1/60 x 20 = 1/3
1/61 + 1/62 + .... + 1/80 > 1/80 + 1/80 + .... + 1/80 = 1/80 x 20 = 1/4
=> 1/41 + 1/42 + .... + 1/80 > 1/3 + 1/4 = 7/2
=> đpcm
vào ccâu hỏi tương tự có dạng jống thế đêý bn
tick cko mik đúng nhé
câu hỏi tương tự ,tick đúng cho mình nhé
Chứng minh rằng :1/41+1/42+1/43+...+1/80 > 7/12
Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12 ( ĐPCM )